Q. 20 - Calculus

Question

Give an example of convergent series $\sum _{k=1}^{\infty }{a}_k$ and $\sum _{k=1}^{\infty }{b}_k$ such that the series $\sum _{k=1}^{\infty }{a}_k{b}_k$diverges.

Step-by-Step Solution

Answer

$\sum _{k=1}^{\infty }{a}_k{b}_k=\sum _{k=1}^{\infty }\left(\frac{1}{k}\right)$ diverges

Step 1. Given information

$\sum _{k=1}^{\infty }{a}_k$ and $\sum _{k=1}^{\infty }{b}_k$ converges

Step 2. Check if ∑ k = 1 ∞ a k and ∑ k = 1 ∞ b k converges

$\sum _{k=1}^{\mathrm{\infty }} (-1{)}^{k+1}{a}_k$and $\sum _{k=1}^{\mathrm{\infty }} (-1{)}^k{a}_k$ both converge

Therefore, 

$\sum _{k=1}^{\mathrm{\infty }} a_k=\frac{(-1{)}^k}{\sqrt{k}}$ converges

Now,

$\sum _{k=1}^{\mathrm{\infty }} b_k=\frac{(-1{)}^k}{\sqrt{k}}$ diverges

Step 3. Check if ∑ k = 1 ∞ a k   b k diverges

$\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }} a_k{b}_k=\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{\sqrt{k}}\cdot \frac{(-1{)}^k}{\sqrt{k}}\\ =\sum _{k=1}^{\mathrm{\infty }} \left(\frac{1}{k}\right)\end{array}$